Page 39 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

1
1.103 If the real numbers a, b, c, d are all distinct, and a, b, c, db + 1 = c + 1 = d + 1 = x
+ =
b c d a
find the value of x .
1
1
1
1
+
+
+
Solution: a, b, c, d = b + 1 = c + 1 = d + 1 = x ⇒ a, b, c, d = x (i), a, b, c, d= x (ii), b, c, d + = x (iii),
a,
b c d a b c d
a,
1 = 1 , substitute it into (ii): b, c, d = x−a , substitute it into (iii):
2
a, b, c, d + = x (iv). (i) implies a, b, c, d x−a x −ax−1
a
x − a 1 3 2
+ = x , that is, dx − (ad + 1)x − (2d − a)x + ad +1 = 0 (v).
2
x − ax − 1 d
(iv) implies ad +1 = ax , substitute it into (v):
3
3
3
dx − ax − 2dx + ax + ax =0 ⇒ (d − a)x − (d − a)2x =0 ⇒ (d − a)(x − 2x)=0.
3
x−a
a,
3
Since d − a �=0, then x − 2x =0. If x =0, then b, c, d = x −ax−1 ⇒ a = c , a
2
√
contradiction. Hence, x − 2= 0 ⇒ x =2 ⇒ x = ± 2.
2
2
1.104 Consider a group of natural numbers a 1 ,a 2 , ··· ,a n , in which there are K i
numbers equal to i (i =1, 2, ··· ,m ). Let S = a 1 + a 2 + ··· + a n ,S j = K 1 + K 2 + ··· + K j
(1 ≤ j ≤ m ). Show S 1 + S 2 + ··· + S m =(m + 1)S m − S .
Proof: S = a 1 + a 2 + ··· + a n = K 1 · 1+ K 2 · 2+ ··· + K n · m =(K 1 + K 2 + ··· + K m )+.
(K 2 + K 3 + ··· + K m )+ ··· + K m = S m + (S m − S 1 )+ ··· + (S m − S m−1 )+ S m − S m =
(m + 1)S m − (S 1 + S 2 + ··· + S m )
Hence, S 1 + S 2 + ··· + S m =(m + 1)S m − S.
1.105 There are ten distinct rational numbers, and the sum of any nine of them is
an irreducible proper fraction whose denominator is 22, find the sum of these ten rational
numbers.

Solution: Let these ten distinct rational numbers be a 1 <a 2 < ··· <a 1 0 . We have
m
(a 1 + a 2 + ··· + a 10 ) − a k = , where k =1, 2, ··· , 10 . m is an odd number and
22
1 ≤ m ≤ 21,m �= 11 . Additionally because a 1 ,a 2 , ··· ,a 10 are all distinct, then
1+3 +5+7+9+13 +15+17 +19+21
10(a 1 +a 2 +···+a 10 )−(a 1 +a 2 +···+a 10 ) = .
22
Hence, a 1 + a 2 + ··· + a 10 =5/9.

1.106 Given a + b + c = abc =0, evaluate
2
2
2
2
2
2
(1 − b )(1 − c ) (1 − a )(1 − c ) (1 − a )(1 − b ) .
+ +
bc ac ab
+b +
Solution: a, b, c, d c = abc �=0 ⇒ ab = a+b+c ⇒ a+b = ab − 1 .
c
c
b + c a + c
Similarly, = bc − 1, = ac − 1 . We have
a b

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